Respuesta :
To solve the problem, it is necessary to use the concepts of gravitational force, centripetal force and trigonometric components that can be extrapolated from the statement.
By definition we know that the Force of Gravity is given by
[tex]F_g=mg[/tex]
Where,
m= Mass
g = Gravitational Acceleration
The centripetal force is given by,
[tex]F_c = \frac{mv^2}{R}[/tex]
Where,
m = Mass
v = Velocity
R = Radius
For the case described in the problem, the Force of gravity the net component would be given by sin?, While for the centripetal force the net component is in the horizontal direction, therefore it corresponds to the [tex]cos\theta[/tex]
Then,
[tex]F_g = mg sin\theta[/tex]
[tex]F_c = \frac{mv^2}{r}cos\theta[/tex]
From the radius we have its length but not the net height, which would be given by
[tex]r = L sin\theta[/tex]
So equating the equations we have to
[tex]F_g = F_c[/tex]
[tex]mg sin\theta=\frac{mv^2}{r}cos\theta[/tex]
[tex]mg sin\theta=\frac{mv^2}{Lsin\theta}cos\theta[/tex]
Re-arrange to find v,
[tex]v = \sqrt{\frac{gLsin^2\theta}{cos\theta}}[/tex]
Replacing with our values
[tex]v = \sqrt{\frac{(9.8)(22*10^{-2})(sin^2 24)}{cos24}}[/tex]
[tex]v = 0.624[/tex]
Therefore the tangential velocity of the mass is 0.624m/s
Tangential speed is the linear component of speed by which an object is moving along a circular path.
Tangential speed of the bob is 0224 m/s.
What is tangential speed?
Tangential speed is the linear component of speed by which an object is moving along a circular path.
Given information-
The mass of the bob is 0.250 kg.
The length of the string is 22.0 cm.
The angle made by the string to the vertical is 24 degrees.
The centripetal force in the vertical direction can be given as,
[tex]F_v=mg\sin \theta[/tex]
The centripetal force in the horizontal direction can be given as,
[tex]F_h=\dfrac{mv^2}{r} \cos \theta[/tex]
As the force in horizontal direction is equal to the force in vertical direction. Thus,
[tex]\begin{aligned}F_v&=F_h\\mg\sin \theta&=\dfrac{mv^2}{r} \cos \theta\\g\sin \theta&=\dfrac{v^2}{r} \cos \theta\\\end[/tex]
Now the radius can be given as,
[tex]r=L\sin \theta[/tex]
Put this value in the above equation as,
[tex]g\sin \theta&=\dfrac{v^2}{L\sin\theta} \cos \theta\\\end[/tex]
Rewrite the equation as,
[tex]v=\sqrt{gL\sin\theta\tan\theta}[/tex]
Put the values in the
[tex]v=\sqrt{9.81\times(22\times10^{-2})\sin({24})\tan({24})}\\v=0.624\rm m/s[/tex]
Hence tangential speed of the bob is 0224 m/s.
Learn more about the tangential speed here;
https://brainly.com/question/4933588
